If radius of the octahedral void is r, radius of the atom in close packing is R, derive relation between r & R.

Explanation

In an hcp or ccp structure of particles, the void that is surrounded by six nearest neighbour particles occupying the eight corners of a regular octahedron is called an octahedral void. Of these six particles, four lie in a plane, two above and two below this plane. A section of an octahedral void in a close-packed structure is shown in the figure.

Suppose, the radius of each of the particles in the close-packed structure = R and the radius of the octahedral void created in this structure = r.
The hypotenuse (BC) of the right-angled triangle ABC,
`BC=sqrt(AB^(2)+AC^(2))=sqrt((R+r)^(2)+(R+r)^(2))`
It is also evident from the figure, BC = 2R
`therefore" "2R=sqrt(2(R+r^(2)))=sqrt2(R+r)`
or, `R(sqrt2-1)=ror,r/R=0.414`
Thus the radius of an octahedral hole in an hcp or ccp structure is equal to 0.414 times the radius of the packed particles.